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Mathematics and Statistics
8-1999
Positive Solutions to Semilinear Problems with
Coefficient that Changes Sign
Nguyen Phuong Cac
Juan A. Gatica
Yi Li
Wright State University - Main Campus, [email protected]
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Repository Citation
Cac, N. P., Gatica, J. A., & Li, Y. (1999). Positive Solutions to Semilinear Problems with Coefficient that Changes Sign. Nonlinear
Analysis: Theory, Methods & Applications, 37 (4), 501-510.
http://corescholar.libraries.wright.edu/math/117
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[email protected].
Positive solutions to semilinear problems with
coefficient that changes sign
Nguyen Phuong Cac, Juan A. Gatica, Yi Li
Department of Mathematics, University of Iowa;
Iowa City IA 52242, USA
1
Introduction
In this paper we will discuss the existence‘and non-existence of positive solutions to the
problem:
−∆u = λa(x)f (u), x ∈ Ω
u(x) = 0, x ∈ ∂Ω.
(1.1)
where Ω will be assumed to be a bounded domain in RN with smooth boundary.
We will concentrate on the case when the coefficient a(x) is allowed to change sign;
this brings about an interesting mathematical problem.
This is not a new problem: due to its appearance in many mathematical models in
physics, several special cases of it have been studied since the middle 1800’s. For example,
it was used by Lord Kelvin [T,W] and J. Lane [L,J] to study the equilibrium configuration
of mass in a spherical cloud gas; this model was further studied by R.H. Fowler [F,R],
leaving the Emden Fowler equation which, in a generalized form is still studied today,
as seen in, for example [AP]. For a review of the work done up to 1975 in the ordinary
differential equations setting, see [W]; for a review of the work done for the existence of
positive solutions of semilinear elliptic equations, see [L].
Positive solutions are usually the ones of interest and because of the difficulties that
this brings in using the techniques developed for nonlinear functional analysis, most of
the recent work assumes nonnegativity of a(x) and f (u) in order to generate positive
operators using the Green’s function of −∆ on the region where the problem is posed.
In nuclear physics, a form of (1.1) was proposed and used by E. Fermi [F,E] and by
L. H. Thomas [T,L], and now it is widely known as the Thomas-Fermi equation.
Both the Emden-Fowler equation as well as the Thomas Fermi equation continue to
be subjects of considerable interest in theoretical physics to this day, as a search on the
Wide World Web will quickly convince you.
In combustion theory equation (1.1) has also been used as a model, for example in
[ACP], [BK],[CL], [G],[L].
Equation (1.1) has also found applications to geometry: see [N]
As mentioned above, most of the literature deals with the problem under the assumption that the coefficient function a(x) be positive in its domain; we are interested in the
case when this function is allowed to change sign. In particular we are interested in determining sufficient conditions on a(x) to assure the existence of positive solutions for
1
small values of λ, to determine ranges of λ for which no positive solution exists, and to
determine conditions under which there are more than one positive solution.
The main motivation for this work are the inherent mathematical difficulties it presents.
We consider the problem of establishing conditions for the existence of positive solutions of equation (1.1).
2
The general case: existence.
In this section we will tackle the general case of existence of positive solutions of (1.1),
imposing as few conditions on Ω as we can and, in particular, we will consider both the
radially symmetric as well as the non-radially symmetric cases.
In another section we will also study the case when positive solutions can be ruled out
for values of λ in certain ranges.
Theorem 2.1. Suppose that f ∈ L∞ (R); a ∈ Ls (Ω), s > 1. Then for every λ ∈ R problem
(1.1) has a solution uλ ∈ W 2,s (Ω).
Proof.
For any v ∈ Ls (Ω) and any µ ∈ [0, 1] let Tµ (v) ∈ W 2,s (Ω) be the solution to the
Dirichlet problem
−∆u = µλa(x)f (v(x)), x ∈ Ω
u(x) = 0, x ∈ ∂Ω.
(2.1)
For some constant K > 0, independent of v, µ, λ, we have:
kTµ (v)k ≤ kTµ (v)kW 2,s ≤ K |λ| kakLs = deg(I − T0 , Bρ (0), 0)
= deg(I, Bρ (0), 0) = 1,
0 ≤ µ ≤ 1.
Thus T1 has a fixed point inside Bρ (0)and this implies that the Dirichlet problem (2.1)
has a solution u ∈ W 2,s (Ω).
Theorem 2.2. Let a ∈ Ls (Ω), s > N/2, and suppose that there exists ε > 0 such that the
solution of the Dirichlet problem
−∆w = a+ (x) − (1 + ε)a− (x), x ∈ Ω
w(x) = 0, x ∈ ∂Ω,
(2.3)
where a+ (x) = max(0, a(x)), a− (x) = a+ (x) − a(x), x ∈ Ω is nonnegative inside Ω.
Suppose further that f ∈ C(R, R), f (0) > 0. Then there exists λ0 > 0 such that for
λ ∈ (0, λ0 ) the Dirichlet problem
−∆u = λa(x)f (u), x ∈ Ω
u = 0 on ∂Ω
has a solution which is positive in Ω.
Proof.
2
Fix a large number M > 0 and define g : R → R as follows:
g(ζ) = f (0), ζ ≤ 0,
g(ζ) = f (ζ), 0 < ζ ≤ M
g(ζ) = f (M ), M < ζ.
By Theorem 1, the Dirichlet problem
−∆u = λa(x)g(u), x ∈ Ω,
u = 0 on ∂Ω
has a solution uλ ∈ W 2,s (Ω). Since s > /2, by Sobolev’s imbedding theorem, implies that
u ∈ C(Ω).
ε
). The continuity of g implies that there is δ ∈ (0, M ) such
Fix a number γ ∈ (0, 2+ε
that
|ζ| < δ ⇒ g(0) − g(0)γ < g(ζ) < g(0) + g(0)γ.
Again the fact that s > N/2, (2.2) and the Sobolev imbedding theorem imply that
there exists λ0 > 0 such that
λ ∈ (0, λ0 ) ⇒ kuλ kC(Ω) < δ.
If we denote by G : Ω × Ω → R the Green’s function for −∆ with homogeneous
boundary conditions, then it is well known that G(x, ζ) > 0, (x, ζ) ∈ Ω × Ω,and, for x ∈ Ω
and λ ∈ (0, λ0 ) :
Z
Z
+
uλ (x) = λ{ G(x, ζ)a (ζ)g(uλ (ζ))dζ − G(x, ζ)a− (ζ)g(uλ (ζ))dζ}
Ω
ZΩ
> λ G(x, ζ){a+ (ζ)[g(0) − g(0)γ] − a− (ζ)[g(0) + g(0)γ]}dζ
Ω
Z
1+γ −
> λf (0)(1 − γ) G(x, ζ)[a+ (ζ) −
a (ζ)]dζ.
1−γ
Ω
From the hypothesis concerning problem (2.3), it follows that:
Z
1+γ
uλ (x) > λf (0)(1 − γ)[(1 + ε) −
] G(x, ζ)a− (ζ)dζ.
1−γ Ω
By our choice of γ the right hand side is nonnegative and thus we have uλ (x) > 0, x ∈
Ω.
Remark 2.3. Theorem 2 improves the result of [CFG] in a number of ways.
First: [CFG] considers only the radially symmetric case.
Second: it uses the method of upper and lower solutions and, as a consequence of
the method, it requires that f be nondecreasing.
Remark 2.4. Third: [CFG] requires that for some τ > 0
3
Z
t
r
0
N −1 +
t
Z
rN −1 a− (r)dr, 0 ≤ t ≤ ν
a (r)dr ≥ (1 + τ )
(2.4)
0
where ν is the radius of the ball Ω = Bν (0).
This condition implies that the radially symmetric solution u(x) = u(kxk) of (1.1) is
positive inside the ball Ω. In fact, since a cannot be zero in Ω, if a− (r) = 0, 0 ≤ r ≤ ν
then, on the one hand, (2.4) holds for any τ > 0; on the other hand for any ε > 0 the
solution of (2.3) is positive inside the ball by the maximum principle. Hence Theorem 2.2
generalizes the main result in [CFG}
So let us consider the case a− not identically zero on [0, ν]. Let ω(x) = ω(kxk) satisfy:
−∆ω = a+ (x) − (1 + τ /2)a− (x), kxk < v
ω(x) = 0, kxk = v.
(2.5)
Then:
[ω(r)rN −1 ]0 = −rN −1 [a+ (r) − (1 + τ /2)a− (r)], 0 ≤ r ≤ v
ω(γ) = 0, ω 0 (0) = 0
(2.6)
Thus:
Rr
Rr
ω 0 (r)rN −1 R= − 0 tN −1 a+ (t)dt + (1 + τ /2) 0 tN −1 a− (t)dt
r
≤ −(τ /2) 0 tN −1 a− (t)dt, 0 ≤ r ≤ v.
(2.7)
Since a− is not identically zero, there exists r0 ∈ (0, ν) such that the right hand side of
1.2.7 is negative on [r0 , ν]. It follows that ω 0 is nonpositive on [0, ν] and negative on [r0 , ν].
Since ω(ν) = 0 it follows that the solution ω is positive inside the ball. Thus hypothesis
(4) of [CFG] is stronger than our hypothesis concerning equation (1.1).
We end this section with a result that holds in our general setting and will be of
considerable importance in what is to come.
Corollary 2.5. If Ω is a ball,f ∈ C(Ω), f (0) > 0, a ∈ Ls (Ω), ṡ > N/2, and if the problem
is radially symmetric, then there exists λ0 ∈ (0, 1) such that problem (2.1) has a positive
solution for λ ∈ (0, λ0 ).
Lemma 2.1. Let a ∈ Ls (Ω), s > N/2, a 6= 0, and consider the problem:
−∆ω = a(x), x ∈ Ω
−∆ω1 = a(x)ω, x ∈ Ω
ω = ω1 = 0 on ∂Ω.
If ∂Ω satisfies Hopf’s boundary lemma, then
∂ω1
<0
∂v
on ∂Ω, where v is the unit outward normal vector.
Proof.
Let W (x) = ω1 (x) − 12 ω 2 (x). Then:
∆W (x) = ∆ω1 (x) − [ω(x)∆ω(x) + |∇ω|2 (x)]
= − |∇ω|2 (x).
4
It follows that W is a strict super harmonic function on Ω unless ∇ω is identically
zero, which would imply that ω is identically zero and thus a must be identically zero.
Thus, under our assumptions, it must be the case that ∆W (x) ≤ 0 and is not identically zero on ∂Ω.
Hopf’s boundary lemma now yields:
0>
3
∂W
∂ω1
∂ω
∂ω1
=
−ω
=
.
∂v
∂v
∂v
∂v
Existence and Non-Existence.
Now we will concentrate our focus in adding conditions which will assure us of much more
detailed information concerning the existence or non-existence of positive solutions for the
problem under study.
The most used hypotheses in this section are listed below. In specific instances we will
add to them.
H1 ) f : R → R is continuous with f (0) > 0.
H2 ) a ∈ Ls (Ω), ṡ > N/2.
H3 ) ΩR is a bounded domain in RN , 0 ∈ Ω, and ∂Ω ∈ C 2 (Ω).
H4 )ε Ω G(x, ζ)(a+ (ζ) − (1 + ε)a− (ζ))dζ > 0, x ∈ Ω.
Rt
H5 ) A(t) = 0 rN −1 a(r)dr ≥ 0 and for some t0 ∈ (0, 1), A(t0 ) > 0.
A direct consequence of Theorem 2 is the following:
Theorem 3.1. Under H1 ) and H4 )ε for some ε > 0,there exists λ0 ∈ (0, 1) such that
problem (1.1) has at least one positive solution for λ ∈ (0, λ0 ).
Now we will prove our first nonexistence result.
Theorem 3.2. If H2 ), H4 )0 hold, if B1 (0) ⊆ Ω and if f 0 (0) < 0 then (1.1) has no bounded
positive in solution in L∞ for λ > 0.
Proof.
Let ρ > 0 and define
b(r) = 1, 0 ≤ r ≤ ρ
= −1, ρ ≤ r ≤ 1.
Define
Z
G(x, ζ)b(ζ)dζ.
ω(x) =
B1 (0)
where G is the Green’s function for −∆ on B1 (0).
Given our assumptions, there exists ρ > 0 so that:
ω(x) > 0, x ∈ B1 (0),
∂ω
= 0 on ∂B1 (0).
∂v
Let u = λW be any solution of (1.1). Then:
−∆W (x) = a(x)f (λw(x)).
5
(3.1)
We have:
Z
G(x, ζ)a(ζ)f (λW (ζ))dζ
W (x) =
B1 (0)
Z
G(x, ζ)a(ζ)[f (0) + f 0 (0)λW (ζ) + o(W (ζ))]dζ
B1 (0)
Z
0
G(x, ζ)a(ζ)W (ζ)[1 + o(1)]dζ
= f (0)ω(x) + λf (0)
=
B1 (0)
0
= f (0)ω(x) + λf (0)ωλ (x),
where
Z
ωλ (x) =
G(x, ζ)a(ζ)W (ζ)[1 + o(1)]dζ.
B1 (0)
As λ → 0+ ,(3.1) becomes ∆W0 + af (0) ≥ 0, concluding that W0 = f (0)ω(x) and
hence
Z
ωλ (x) →
G(xζ)a(ζ)ω(ζ)f (0)dζ = ω2 (x).
B1 (0)
Here we have
−∆ω2 (x) = f (0)a(x)ω(x) on B1 (0),
∂ω2
< 0 on ∂B1 (0).
∂v
Hence
∂ωλ
<0
∂v
for λ sufficiently small. For these values of λ we get:
∂W
∂ω
∂ωλ
= f (0)
+ λf 0 (0)
>0
∂v
∂v
∂v
and this together with the fact that W = 0 on ∂B1 (0) implies that W is negative near
the boundary.
Remark 3.3. The same argument implies that if f 0 (0) > 0 and
W is positive for λ small and positive.
∂ω2
∂v
< 0 on ∂B1 (0) then
An immediate consequence is:
Theorem 3.4. If H2 ) and H4 )0 hold, if Ω = B1 (0) and if f 0 (0) > 0 then there is λ0 ∈ (0, 1)
such that problem (1.1) has a positive solution for λ ∈ (0, λ0 ).
Theorem 3.5. Assume that the Ω = B1 (0), that the problem is radially symmetric, that
a : [0, 1] → R is nonincreasing, that f : R → R is locally Lipschitz and positive and such
that there exists a positive nondecreasingR function g with g(u) = mins≥u f (s) satisfies
∞ du
< ∞ and there exists ε > 0 such that
the strictly superlinear growth condition 0 g(u)
a(x) > ε in Bε (0).
Then the problem
has no solution if λ ≥
2N
ε3
−∆u = λaf (u) in B1 (0),
u > 0 in B1 (0)
u = 0 on ∂B1 (0)
R ∞ du
.
0 g(u)
6
Proof.
If u is a positive solution, then by [GNN], u is radially symmetric and so:
u00 +
N −1 0
u + λa(r)f (u) = 0.
r
In Bε (0) we then have:
r
Z
N −1 0
r
sN −1 a(s)f (u(s))ds = 0.
u (r) + λ
0
This implies that
r
Z
N −1 0
r
sN −1 g(u(s)ds ≤ 0.
u (r) + λε
0
0
Since u < 0 in (0, ε), we have that:
r
r
Z
N −1 0
sN −1 ds ≤ 0.
u (r) + λεg(u(r))
0
We get:
Thus:
u0 (r) λε
+ r ≤ 0.
g(u)
N
Z
0
This shows that:
ε
u0 (r)
dr +
g(u(r))
λε3
≤
2N
Z
u(0)
u(ε)
Z
ε
0
λε
rdr ≤ 0.
N
du
≤
g(u)
Z
∞
0
du
.
g(u)
This yields the desired result.
Theorem 3.6. If there exists ε > 0 such that Bε (x0 ) ⊆R Ω,if a(x) ≥ ε on Bε (x0 ) and if
∞ du
f (u) ≥ g (u) ≥ 0, where g is nondecreasing, convex,with 0 g(u)
< ∞, then the problem
−∆u = λa(x)f (u) in Bε (x0 )
u > 0 in Bε (x0 )
has no solution if
2N
λ≥ 4
ε
Z
0
∞
du
.
g(u)
Proof.
R
Let u(r) = ∂Br (x0 ) u(x)dSx for 0 < r < ε.
In Bε (x0 ) we have:
Z
[∆u + af (u)] = 0
∂Br (x0 )
and hence:
Z
[∆u + λεg(u)] ≤ 0
Br (x0 )
in Bε (x0 ).
7
Jensen’s inequality implies that:
∆u + λε2 g(u) ≤ 0.
Similarly we get:
2N
λ< 4
ε
4
Z
0
∞
du
.
g(u)
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8
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9